Integrate. $ \int 8\cos(x)\,dx $ $=$ $+ C$
We need a function whose derivative is $8\cos(x)$. We know that the derivative of $\sin(x)$ is $\cos(x)$, so let's start there: $\dfrac{d}{dx} \sin(x) = \cos(x)$ Now let's multiply by $8$ : $\dfrac{d}{dx}\left[ 8\sin(x) \right]= 8\cos(x)$ Because finding the integral is the opposite of taking the derivative, this means that: $ \int 8\cos(x)\,dx =8 \sin(x)\, + C$ The answer: $8\sin(x)+C$.